Question

A ring of radius $R$ is rotating with an angular speed ${\omega }_0$ about a horizontal axis. It is placed on a rough horizontal table. The coefficient of kinetic friction is ${\mu }_k$. The time after which it starts rolling is:

• A. $\frac{{\omega }_0{\mu }_{k}R}{2g}$
• B. $\frac{{\omega }_{0}R}{2{\mu }_{k}R}$
• C. $\frac{2{\omega }_{0}R}{{\mu }_{k}g}$
• D. $\frac{{\omega }_{0}R}{2{\mu }_{k}g}$

Answer 1

Answer: (D)

Acceleration produced in the centre of mass due to friction
$a=\frac{f}{M}=\frac{{\mu }_{k}Mg}{M}={\mu }_{k}g\dots (i)$
Angular retardation produced by the torque due to friction
$\alpha =\frac{\tau }{I}=\frac{fR}{I}=\frac{{\mu }_{k}MgR}{I}=\frac{{\mu }_{k}g}{R}\dots (ii)$
$\therefore v=0+{\mu }_{k}gt(\therefore u=0)$ (Using (i))
$\therefore \omega ={\omega }_0-\frac{{\mu }_{k}gt}{R}$ ( Using (ii))
For rolling without slipping
$v=R\omega$
$\frac{{\mu }_{k}gt}{R}={\omega }_0-\frac{{\mu }_{k}gt}{R}$
$\Rightarrow t=\frac{R{\omega }_0}{2{\mu }_{k}g}$